Let `f_(1) (x) and f_(2) (x)` be twice differentiable functions where `F(x)= f_(1) (x) + f_(2) (x) and G(x) = f_(1)(x) - f_(2)(x), AA x in R, f_(1) (0) = 2 and f_(2)(0)=1. "If" f'_(1)(x) = f_(2) (x) and f'_(2) (x) = f_(1) (x) , AA x in R` then the number of solutions of the equation `(F(x))^(2) =(9x^(4))/(G(x))`is...... .

Answer

To solve the problem, we need to analyze the given functions and the equation step by step.

### Step 1: Define the Functions
We are given two functions:
- \( F(x) = f_1(x) + f_2(x) \)
- \( G(x) = f_1(x) - f_2(x) \)

### Step 2: Differentiate the Functions
We differentiate \( F(x) \) and \( G(x) \):
- \( F'(x) = f_1'(x) + f_2'(x) \)
- \( G'(x) = f_1'(x) - f_2'(x) \)

### Step 3: Use the Given Derivatives
From the problem, we know:
- \( f_1'(x) = f_2(x) \)
- \( f_2'(x) = f_1(x) \)

Substituting these into the derivatives:
- \( F'(x) = f_2(x) + f_1(x) = F(x) \)
- \( G'(x) = f_2(x) - f_1(x) = -G(x) \)

### Step 4: Solve the Differential Equations
We have:
1. \( F'(x) = F(x) \)
2. \( G'(x) = -G(x) \)

These are standard first-order differential equations.

For \( F(x) \):
- The solution is \( F(x) = A e^x \) for some constant \( A \).

For \( G(x) \):
- The solution is \( G(x) = B e^{-x} \) for some constant \( B \).

### Step 5: Use Initial Conditions
We know:
- \( f_1(0) = 2 \)
- \( f_2(0) = 1 \)

Thus:
- \( F(0) = f_1(0) + f_2(0) = 2 + 1 = 3 \)
- \( G(0) = f_1(0) - f_2(0) = 2 - 1 = 1 \)

Substituting these into our solutions:
1. \( F(0) = A e^0 = A = 3 \)
2. \( G(0) = B e^0 = B = 1 \)

So we have:
- \( F(x) = 3 e^x \)
- \( G(x) = e^{-x} \)

### Step 6: Substitute into the Given Equation
We need to solve the equation:
\[ (F(x))^2 = \frac{9x^4}{G(x)} \]

Substituting \( F(x) \) and \( G(x) \):
\[ (3 e^x)^2 = \frac{9x^4}{e^{-x}} \]
This simplifies to:
\[ 9 e^{2x} = 9x^4 e^x \]
Dividing both sides by 9:
\[ e^{2x} = x^4 e^x \]
Rearranging gives:
\[ e^{2x - x} = x^4 \quad \Rightarrow \quad e^x = x^4 \]

### Step 7: Analyze the Equation
We need to find the number of solutions to \( e^x = x^4 \).

### Step 8: Graphical Analysis
- The function \( e^x \) is an exponential function that increases rapidly.
- The function \( x^4 \) is a polynomial function that also increases but at a slower rate for large \( x \).

### Step 9: Intersection Points
To find the number of solutions, we can analyze the graphs:
1. At \( x = 0 \), \( e^0 = 1 \) and \( 0^4 = 0 \) (1 intersection).
2. As \( x \) increases, \( e^x \) grows faster than \( x^4 \).
3. For negative \( x \), \( e^x \) approaches 0 while \( x^4 \) approaches 0 as well.

By plotting the graphs, we can see that they intersect at two points.

### Conclusion
Thus, the number of solutions to the equation \( (F(x))^2 = \frac{9x^4}{G(x)} \) is **2**.

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